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题意：吕布有n个武将，曹操有m（m>=n）个武将。给出k个关系，为吕布的某个武将和曹操的某个武将pk后会受到的伤害。吕布要求他所有n的武将都要上场，每个武将只能战斗一次，问如何安排，使得所有武将受到的伤害总和最小。
思路：裸的KM。 用map把武将名字变成点的编号。
1/* ***********************************************
2Author :111qqz
3Created Time :2016年06月02日 星期四 19时17分19秒
4File Name :code/hdu/2813.cpp
5************************************************ */
6
7#include <cstdio>
8#include <cstring>
9#include <iostream>
10#include <algorithm>
11#include <vector>
12#include <queue>
13#include <set>
14#include <map>
15#include <string>
16#include <cmath>
17#include <cstdlib>
18#include <ctime>
19#define fst first
20#define sec second
21#define lson l,m,rt<<1
22#define rson m+1,r,rt<<1|1
23#define ms(a,x) memset(a,x,sizeof(a))
24typedef long long LL;
25#define pi pair < int ,int >
26#define MP make_pair
27
28using namespace std;
29const double eps = 1E-8;
30const int dx4[4]={1,0,0,-1};
31const int dy4[4]={0,-1,1,0};
32const int inf = 0x3f3f3f3f;
33const int N=205;
34int n,m,k;
35map<string,int>mp,mp2;
36int tot1,tot2;
37int w[N][N];
38int lx[N],ly[N];
39int link[N];
40bool visx[N],visy[N];
41int slk[N];
42
43
44bool find( int u)
45{
46 visx[u] = true;
47 for ( int v = 1 ; v <= m ; v++)
48 {
49 if (visy[v]) continue;
50 int tmp = lx[u] + ly[v] - w[u][v];
51 if (tmp==0)
52 {
53 visy[v] = true;
54 if (link[v]==-1||find(link[v]))
55 {
56 link[v] = u;
57 return true;
58 }
59 }else if (tmp<slk[v]) slk[v] = tmp;
60 }
61 return false;
62}
63int KM()
64{
65 ms(lx,0xc0);
66 ms(ly,0);
67 ms(link,-1);
68
69 for ( int i = 1 ; i <= n ; i++)
70 for ( int j = 1 ; j <= m ; j++)
71 lx[i] = max(lx[i],w[i][j]);
72
73 // for ( int i = 1 ; i <= n ; i++) cout<<"lx[i]:"<<lx[i]<<endl;
74 for ( int i = 1; i <= n ; i++)
75 {
76 ms(slk,0x3f);
77
78 while (1)
79 {
80 ms(visx,false);
81 ms(visy,false);
82
83 if (find(i)) break;
84
85 int d = inf;
86
87 for ( int j = 1 ; j <= m ; j++)
88 if (!visy[j]&&slk[j]<d) d = slk[j];
89
90 for ( int j = 1 ; j <= n ; j++) if (visx[j]) lx[j]-=d;
91 for ( int j = 1 ; j <= m ; j++) if (visy[j]) ly[j]+=d; else slk[j]-=d;
92 }
93 }
94 int res = 0;
95 for ( int i = 1 ; i <= m ; i++)
96 if (link[i]>-1) res += w[link[i]][i];
97
98 return -res;
99}
100int main()
101{
102 #ifndef ONLINE_JUDGE
103 freopen("code/in.txt","r",stdin);
104 #endif
105// ios::sync_with_stdio(false);
106 while (scanf("%d%d%d",&n,&m,&k)!=EOF)
107 {
108 mp.clear();
109 mp2.clear();
110 tot1 = tot2 = 0;
111 ms(w,0xc0);
112 string u,v;
113 char tmpu[25],tmpv[25];
114 int cost;
115 for ( int i = 1 ; i <= k ; i++)
116 {
117 //cin>>u>>v>>cost;
118 scanf("%s %s %d",tmpu,tmpv,&cost);
119 u = string(tmpu);
120 v = string(tmpv);
121 if (!mp[u]) mp[u] = ++tot1;
122 if (!mp2[v]) mp2[v] = ++tot2;
123 w[mp[u]][mp2[v]] = -cost;
124 }
125
126 int ans = KM();
127 //cout<<ans<<endl;
128 printf("%d\n",ans);
129 }
130
131 #ifndef ONLINE_JUDGE
132 fclose(stdin);
133 #endif
134 return 0;
135}